Comparing 2 Improper Integrals: Convergence & Criteria

[DavideGenoa]

I read that the improper Riemann integral $\int_0^1 \frac{1}{x}\sin\frac{1}{x}dx$ converges and that $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx$ does not.
I have tried comparison criteria for $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx$, but I cannot find a function $f$ with a divergent integral such that $0\leq f(x)\leq|\frac{1}{x}\sin\frac{1}{x}|$.
I heartily thank you for any help!

EDIT: I have managed to use complex analysis to show that $\int_0^{\infty} \frac{1}{x}\sin xdx=\frac{\pi}{2}$, but, with a change of variable, I see that $\int_0^1 \frac{1}{x}\sin\frac{1}{x}dx=\int_1^{\infty}\frac{1}{x}\sin x dx$. $\frac{1}{x}\sin x$ is integrable on $[0,1]$ (it can be "made continuous" in 0 where it approaches 1), therefore $\int_0^1 \frac{1}{x}\sin\frac{1}{x}dx$ converges.

 

[mfb]

A lower bound for the integrand is "f(x)=0.1/x if |sin(1/x)| > 0.1, f(x)=0 otherwise". The first case always has larger intervals than the second, so you can combine those intervals to find another lower bound that diverges.

 

[mathman]

Let y = 1/x, for both. You will see that the absolute value case can be easily studied. |sinx/x| < c/x for most of the domain of integration.

 

[DavideGenoa]

Thank you both!

|sinx/x| < c/x for most of the domain of integration.

Forgive my ununderstanding: how can we use the fact to show that $\int_{0}^{1}\frac{1}{x}|\sin(\frac{1}{x}|) dx$, which I do see to be the same as $\int_{1}^{+\infty}\frac{1}{x}|\sin(x)| dx$, diverges?

A lower bound for the integrand is "f(x)=0.1/x if |sin(1/x)| > 0.1, f(x)=0 otherwise". The first case always has larger intervals than the second, so you can combine those intervals to find another lower bound that diverges.

Excuse me: how do we rigourously see that the intervals where $f(x)=0.1/x$ are larger than where $f(x)=0$, and how do we see that it diverges? I do know that $\frac{1}{10}\int_{0}^{1}\frac{1}{x}dx$ does, but "some pieces" are lacking here and I am not having any idea to verify that...

I heartily thank you both again!

 

[mathman]

Thank you both!
Forgive my ununderstanding: how can we use the fact to show that $\int_{0}^{1}\frac{1}{x}|\sin(\frac{1}{x}|) dx$, which I do see to be the same as $\int_{1}^{+\infty}\frac{1}{x}|\sin(x)| dx$, diverges?

Excuse me: how do we rigourously see that the intervals where $f(x)=0.1/x$ are larger than where $f(x)=0$, and how do we see that it diverges? I do know that $\frac{1}{10}\int_{0}^{1}\frac{1}{x}dx$ does, but "some pieces" are lacking here and I am not having any idea to verify that...

I heartily thank you both again!

All you need to check is one cycle of sinx.

 

[mfb]

One half-cycle will do the job. If you want to do it in detail, it will take some steps, but it is possible. The value of 0.1 is chosen in a way to give enough room for useful approximations (but you can also use 0.01 if you want).